Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $2.6$ years; the standard deviation is $0.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living between $1.4$ and $3.2$ years.
$2.6$ $2$ $3.2$ $1.4$ $3.8$ $0.8$ $4.4$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $2.6$ years. We know the standard deviation is $0.6$ years, so one standard deviation below the mean is $2$ years and one standard deviation above the mean is $3.2$ years. Two standard deviations below the mean is $1.4$ years and two standard deviations above the mean is $3.8$ years. Three standard deviations below the mean is $0.8$ years and three standard deviations above the mean is $4.4$ years. We are interested in the probability of a lizard living between $1.4$ and $3.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the lizards will have lifespans within 1 standard deviation of the mean. The probability of a particular lizard living between $1.4$ and $3.2$ years is $\color{orange}{13.5\%} + {68\%}$, or $81.5\%$.